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\title{Functionals}
\author{Roland Becker}
\date{\today}

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\section{Definition of functionals}\label{sec:}
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Suppose that $u\in u_{0}+V$ solves the equation
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\begin{equation}\label{eq:state}
a(u)(v) = l(v)\quad\forall v\in V.
\end{equation}
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This corresponds to the strong implementation of Dirichlet boundary conditions in \textsl{cg} or \textsl{nc}.

A functional is defined as
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\begin{equation}\label{eq:}
J(u) := \int_{\Omega} \xi\,dx + \int_{\partial\Omega}\zeta\,dx + \left( a(u)(w)-l(w)\right)
=:J_{1}(u) + J_{2}(u)+J_{3}(u).
\end{equation}
%
where $w|_{\Gamma} = \eta$. The three parts of the functional are defined by
\begin{enumerate}
\item \courier{RightHandSideInterface} ($\xi$)
\item \courier{NeumannInterface} ($\zeta$)
\item \courier{DirichletInterface} ($\eta$)
\end{enumerate}

La signification des deux premières termes est évidente. Pour la troisième, on remarque d'abord qu'elle ne dépend que de 
$\eta$ : soient $w_{1}$ et $w_{2}$ tels que $w_{i}|_{\Gamma}=\eta$, $i=1,2$. Alors on a à cause de (\ref{eq:state})
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\begin{eqnarray*}
  a(u)(w_{1}) -l(w_{1})=  a(u)(w_{1}) - l(w_{1}) +  a(u)(w_{2}-w_{1}) - l(w_{2}-w_{1}) -  =  a(u)(w_{2}) -  l(w_{2}). 
\end{eqnarray*}
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Pour comprendre la signification de cette fonctionnelle, on suppose que $l(v)=\int_{\Omega}f\,dx$ et
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\begin{eqnarray*}
a(u)(v) = \int_{\Omega} k(u)\nabla u\cdot \nabla v\,dx.
\end{eqnarray*}
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On a alors 
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\begin{eqnarray*}
 a(u)(w) -l(w) &=& -\int_{\Omega} \left(\div(k(u)\nabla u)+ f\right)w\,dx + \int_{\partial\Omega} k(u) \frac{\partial u}{\partial n}w\,ds\\
 &=& \int_{\partial\Omega} k(u) \frac{\partial u}{\partial n}\eta\,ds
\end{eqnarray*}
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Si $\eta_{\Gamma}=1$ et $\eta_{\Omega\setminus\Gamma}=0$ on obtient alors le flux moyen sur $\Gamma$:
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\begin{equation}\label{eq:}
a(u)(w) -l(w) =\int_{\partial\Gamma} k(u) \frac{\partial u}{\partial n}\,ds.
\end{equation}

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\subsection{Analyse d'erreur pour $J_{3}$}\label{subsec:}
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Soit $V_{h}$ un espace d'approximation de $V$. Nous avons alors avec $\tilde a(u,u_{h})(v,w):=\int_{0}^{1}a'(u_{h}+s(u-u_{h}))(v,w)\,ds$
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\begin{equation}\label{eq:}
\tilde a(u,u_{h})(u-u_{h},v_{h}) = 0 \quad\forall v_{h}\in V_{h}.
\end{equation}
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Let $z$ satisfy $z|_{\partial\Omega}=\eta$ and
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\begin{equation}\label{eq:}
a'(u)(v,z) = 0\quad\forall v\in V.
\end{equation}
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In addition let $z_{h}|_{\partial\Omega}=\eta_{h}$ and
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\begin{equation}\label{eq:}
a'(u_{h})(v_{h},z_{h}) = 0\quad\forall v_{h}\in V_{h}.
\end{equation}
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It follows since $I_{h}z-z_{h}\in V_{h}$
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\begin{eqnarray*}
\int_{\partial\Omega} k(u) \frac{\partial u}{\partial n}\eta\,ds - \left(a(u_{h})(z_{h}) -l(z_{h}) \right) &=&
\left(a(u)(z) -l(z)\right)- \left(a(u_{h})(z_{h}) -l(z_{h}) \right)
\end{eqnarray*}
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\subsection{Definition of $J_{3}$ in the case of weak boundary conditions}\label{subsec:}
%-------------------------------------------------------------------------
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We suppose that $u\in V$ satisfies
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\begin{eqnarray*}
\begin{split}
\int_{\Omega} k(u)\nabla u\cdot \nabla v\,dx - \int_{\partial\Omega}k(u)\frac{\partial u}{\partial n}v\,ds - \int_{\partial\Omega}k(u)u\frac{\partial v}{\partial n}\,ds+\frac{\gamma}{h}\int_{\partial\Omega}k(u)uv\,ds\\=
\int_{\Omega}fv\,dx - \int_{\partial\Omega}k(u)g\frac{\partial v}{\partial n}\,ds+\frac{\gamma}{h}\int_{\partial\Omega}k(u)gv\,ds.
\end{split}
\end{eqnarray*}



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